• question_answer The angle between the lines joining the points of intersection of line $y=3x+2$ and the curve ${{x}^{2}}+2xy+3{{y}^{2}}+4x+8y-11=0$ to the origin, is A)            ${{\tan }^{-1}}\left( \frac{3}{2\sqrt{2}} \right)$       B)            ${{\tan }^{-1}}\left( \frac{2}{2\sqrt{2}} \right)$ C)            ${{\tan }^{-1}}\left( \sqrt{3} \right)$                            D)            ${{\tan }^{-1}}\left( \frac{2}{2\sqrt{2}} \right)$

Finding the equation of lines represented by the points of intersection of curve and line with origin, we get ${{x}^{2}}+2xy+3{{y}^{2}}+(4x+8y)\left( \frac{y-3x}{2} \right)-11{{\left( \frac{y-3x}{2} \right)}^{2}}=0$            $\Rightarrow {{x}^{2}}+2xy+3{{y}^{2}}+(2xy-6{{x}^{2}}+4{{y}^{2}}-12xy)$                                                 $-\frac{11}{4}{{y}^{2}}-\frac{99}{4}{{x}^{2}}+\frac{33}{2}xy=0$            Proceed and find the angle between the lines represented by it using$\alpha ={{\tan }^{-1}}\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}$.